# Restriction of Mapping is Mapping/General Result

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## Theorem

Let $f: S \to T$ be a mapping.

Let $X \subseteq S$.

Let $f \sqbrk X \subseteq Y \subseteq T$.

Let $f \restriction_{X \times Y}$ be the restriction of $f$ to $X \times Y$.

Then $f \restriction_{X \times Y}: X \to Y$ is a mapping:

## Proof

### $f \restriction_{X \times Y}$ is Many-to-one

We have:

\(\ds \forall x \in X: \, \) | \(\ds \tuple {x, y_1}, \tuple {x, y_2} \in f \restriction_{X \times Y}\) | \(\leadsto\) | \(\ds \tuple {x, y_1}, \tuple {x, y_2} \in f \cap \paren{X \times Y}\) | Definition of Restriction of Mapping | ||||||||||

\(\ds \) | \(\leadsto\) | \(\ds \tuple {x, y_1}, \tuple {x, y_2} \in f\) | Intersection is Subset | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds y_1 = y_2\) | Definition of Mapping |

$\Box$

### $f \restriction_{X \times Y}$ is Left-total

We have:

\(\ds \forall x \in X\) | \(\leadsto\) | \(\ds x \in S\) | Definition of Subset | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds \exists y \in T: \tuple {x, y} \in f\) | Definition of Mapping | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds \exists y \in T: y \in f \sqbrk X\) | Definition of Image of subset | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds \exists y \in Y: \tuple {x, y} \in f\) | Subset Relation is Transitive | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds \exists y \in Y: \tuple {x, y} \in f \cap \paren{X \times Y}\) | Definition of Cartesian product and Definition of Set intersection | |||||||||||

\(\ds \) | \(\leadsto\) | \(\ds \exists y \in Y: \tuple {x, y} \in f \restriction_{X \times Y}\) | Definition of Restriction of mapping |

$\Box$

Hence by definition, $f \restriction_{X \times Y}: X \to Y$ is a mapping.

By definition of domain, the domain of $f \restriction_{X \times Y}$ is $X$.

By definition of codomain, the codomain of $f \restriction_{X \times Y}$ is $Y$.

From Preimage of Mapping equals Domain, the preimage of $f \restriction_{X \times Y}$ is also $X$.

$\blacksquare$